cf #363 b

B. One Bomb

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").

You have one bomb. If you lay the bomb at the cell (x, y), then after triggering it will wipe out all walls in the row x and all walls in the column y.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and columns in the depot field.

The next n lines contain m symbols "." and "*" each — the description of the field. j-th symbol in i-th of them stands for cell (i, j). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.

Output

If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).

Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

Examples

input

3 4
.*..
....
.*..

output

YES
1 2

input

3 3
..*
.*.
*..

output

NO

input

6 5
..*..
..*..
*****
..*..
..*..
..*..

output

YES
3 3

#include<stdio.h>

#include<iostream>

#include<map>

using namespace std;

char ma[][];

int main(){

    int n,m;

    int x=-;

    int y=-;

    int hang[];

    int lie[];

    int cnt=;

    scanf("%d%d",&m,&n);

    for(int i=;i<m;i++) hang[i]=;

    for(int j=;j<n;j++) lie[j]=;

    for(int i=;i<m;i++){

        scanf("%s",ma[i]);

    }

    for(int i=;i<m;i++){

        for(int j=;j<n;j++){

            if(ma[i][j]=='*'){

                hang[i]++;

                lie[j]++;

                cnt++;

            }

        }

    }

    for(int i=;i<m;i++){

        for(int j=;j<n;j++){

            int tmp=hang[i]+lie[j];

            if(ma[i][j]=='*') tmp--;

            if(tmp==cnt){

                printf("YES\n%d %d",i+,j+);

                return ;

            }

        }

    }

    printf("NO\n");

    return ;

}

分别统计行和列中到墙的个数，如果某个行和列中包含全部到墙，则为答案。

巴特西

cf #363 b

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